2017 Multi-University Training Contest 1 hdu 6034

HDU 6034

思路：

AC代码：

#include "iostream"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#define ll long long
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a) memset(a,0,sizeof(a))
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
using namespace std;
const long long INF = 1e18+1LL;
const int inf = 1e9+1e8;
const int N=1e5+100;
const ll mod=1e9+7;

///1002
char s[N];
int lm,n;
ll mi[N];
map<char,int> M;
struct Node{
    int a[N],id;
    bool friend operator< (Node x, Node y){
        for(int i=lm; i>=1; --i){
            if(x.a[i]!=y.a[i]) return x.a[i]>y.a[i];
        }
        return 0;
    }
};
Node bit[26];
int main(){
    int t=1;
    for(ll i=0,base=1;i<N;i++,base=(base*26)%mod) mi[i] = base;
    //ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    while(~scanf("%d", &n)){
        M.clear(),lm=0;
        for(int i=0; i<=25; ++i) bit[i].id=200;
        while(n--){
            scanf("%s", s);
            int ls=strlen(s),b=ls;
            if(ls>1) M[s[0]-'a']=1;
            lm=max(lm,ls);
            for(int i=0; i<ls; ++i){
                bit[s[i]-'a'].a[b-i]++;
                bit[s[i]-'a'].id=s[i]-'a';
            }
        }
        for(int i=0; i<=25; ++i){
            for(int j=1;j<=N-1; ++j){
                if(bit[i].a[j]>25){
                    bit[i].a[j+1]+=bit[i].a[j]/26;
                    bit[i].a[j]%=26;
                    lm=max(lm,j+1);
                }
            }
        }
        sort(bit,bit+26);
        ll r=26; int p=-1;
        for(int i=0; i<=25; ++i){
            if(bit[i].id==200) break;
            --r;
        }//for(int i=0; i<=25; ++i) cout<<mp[i]<<" ";
        if(r==0){
            for(int i=25; i>=0; --i){
                if(!M[bit[i].id]){
                    p=bit[i].id;
                    break;
                }
            }
        } //for(int i=0; i<=25; ++i) cout<<mp[i]<<" ";
        ll ans=0,f; r=26;
        for(int i=0; i<=25; ++i){
            if(bit[i].id==p) f=0;
            else f=--r;
            for(int j=1; j<=lm; ++j){
                ans+=(((f*bit[i].a[j])%mod)*mi[j-1])%mod;
                bit[i].a[j]=0;
            }
        }
        printf("Case #%d: %lld
", t++, (ans+mod)%mod);
    }
    return 0;
}